Question

A particle is moving eastward with a speed of 5 metre per second in 10 second the direction changes towards north but speed remains the same the average acceleration in this time is

Answer 1

Answer:

Initial velocity, u = 5 m/s due east

Final velocity, v = 5 m/s due north

Change in velocity, Δv = v – u = 5 m/s due north - 5 m/s due east

=> Δv = 5 m/s due north + 5 m/s due west [negative of east direction is west]

Δv = (v2 + u2)1/2

=> Δv = 5√2 m/s

Direction of Δv is given by,

tanθ = v/u = -1

=> θ = -45o

Thus, the acceleration is a = Δv/t = 5√2/10 = 1/√2 m/s2 and is directed towards North-West direction

Explanation:

may this help you

Answer 2

Explanation:

acc. = v-u/t

avg acc = ∆v/∆t

here u=0m/s2 and v= 5m/s2

hence ∆v = 5-0 =5 ∆t= 10-0

avg acc= ∆v/∆t= 5/10= 1/2=0.5 m/s2